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Worked Examples · Example 1

Q.A rectangular parallelepiped (box) is drawn with its edges parallel to the coordinate axes, one vertex at the origin O(0,0,0)O(0,0,0) and the diagonally opposite vertex at P(2,4,5)P(2,4,5); the three edges from OO run along the xx-, yy- and zz-axes. Let FF be the vertex of this box that is the foot of the perpendicular dropped from PP onto the XZ-plane (the plane y=0y=0) — i.e. the corner having the same xx- and zz-coordinates as PP but lying in the plane y=0y=0. Find the coordinates of FF.

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✓ Free question

The point FF is where the perpendicular from P(2,4,5)P(2,4,5) meets the XZ-plane. Projecting onto that plane keeps the xx- and zz-coordinates and makes the yy-coordinate zero, giving F=(2,0,5)F=(2,0,5).

Concept

When a rectangular box has one vertex at the origin and the opposite vertex at P(2,4,5)P(2,4,5) with edges along the axes, each of the other vertices is obtained by moving PP back along one or more axes until it meets a coordinate plane. A point lies in the XZ-plane exactly when its distance measured along the yy-axis (OYOY) is zero, i.e. when its yy-coordinate equals 00.

Why this works

The foot of the perpendicular from any point (x,y,z)(x,y,z) onto the XZ-plane is (x,0,z)(x,0,z): the perpendicular from the point to the plane y=0y=0 runs parallel to the yy-axis, so only the yy-coordinate changes (to 00), while xx and zz are preserved.

Steps

  1. Coordinates of PP: x=2, y=4, z=5x=2,\ y=4,\ z=5.
  2. FF lies in the XZ-plane, so its yy-coordinate is 00.
  3. Since FF shares the same xx and zz as PP: xF=2x_F=2 and zF=5z_F=5.
  4. Therefore F=(2, 0, 5)F=(2,\,0,\,5).
✓Final answer

F=(2, 0, 5)F=(2,\,0,\,5).

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